信息学奥赛一本通T1312-昆虫繁殖-题解（各语言代码）

每对成虫过x个月产y对卵，每对卵过两个月长成成虫
设第n个月有a[n]只成虫，b[n]只卵，可列出递推式
\begin{cases}a[n]=a[n-1]+b[n-2]\\b[n]=y\times a[n-x]\end{cases}
既然不需要卵的数量，将两式合并，得到成虫数量公式
a[n]=a[n-1]+y\times a[n-x-2]
递推求解即可
python

x,y,z=map(int,input().split())
l=[1]*(x+2)
for i in range(z+~x):l.append(l[-1]+y*l[i])
print(l[z])

C++

#include<deque>
#include<iostream>
using namespace std;
int main(){
    size_t x,y,z;
    cin>>x>>y>>z;
    deque<size_t>d(x+2,1);
    for(size_t i=z-x;--i;d.pop_front())
        d.push_back(d.back()+y*d.front());
    cout<<d.back();
}

java

import java.util.Scanner;
import java.util.LinkedList;
import java.util.stream.IntStream;
import java.util.stream.Collectors;
public class Main {
    public static void main(String[] args) {
        final Scanner in = new Scanner(System.in);
        final int x = in.nextInt(), y = in.nextInt(), z = in.nextInt();
        LinkedList<Long> l = IntStream.range(0, x + 2).mapToObj(i -> 1L).collect(Collectors.toCollection(LinkedList::new));
        IntStream.range(0, z+~x).forEach(i -> l.add(l.getLast() + y * l.poll()));
        System.out.print(l.getLast());
    }
}